Question

A force \( F \) is applied upon a square with side \( L \). Errors in \( L \) and \( F \) are 2% and 4% respectively. Error in measuring pressure will be

• A. 4 percent
• B. 6 percent
• C. 2 percent
• D. 8 percent

Hint:

When calculating the percentage error in derived quantities, remember to add the percentage errors for multiplication and apply the appropriate exponent for powers (e.g., \( L^2 \)).

Answer 1

The Correct Option is B

Step 1: Understanding the pressure formula.
Pressure \( P \) is given by the formula: \[ P = \frac{F}{A} = \frac{F}{L^2} \] where \( A = L^2 \) is the area of the square. We need to find the error in \( P \) due to errors in \( F \) and \( L \). Step 2: Calculating the percentage error.
The percentage error in pressure \( \delta P \) is given by the sum of the percentage errors in \( F \) and \( L^2 \), using the following formula: \[ \delta P = \delta F + 2 \delta L \] where: - \( \delta F = 4% \) (given error in \( F \)), - \( \delta L = 2% \) (given error in \( L \)). Step 3: Substituting values.
Substitute the values of \( \delta F \) and \( \delta L \) into the formula: \[ \delta P = 4% + 2 \times 2% = 4% + 4% = 6% \] Step 4: Conclusion.
The error in measuring pressure is \( \boxed{6%} \). Thus, the correct answer is (2).

Answer 2

Pressure is defined here as \( P = \dfrac{F}{A} \), where the area \( A \) of the square depends on its side \( L \). Since pressure is obtained by dividing one measured quantity by another, the overall percentage error in \( P \) is found by combining the percentage errors of the quantities that go into it.

1. 4 percent: This is only the error contributed by the force \( F \) on its own. Since the area also carries a measurement uncertainty (through \( L \)), the pressure's total error must include that contribution too, so 4% alone is incomplete.
2. 6 percent: Combining the 4% uncertainty from the force measurement with the 2% uncertainty carried through from the side length's contribution to the area gives a total percentage error of \( 4\% + 2\% = 6\% \) for the pressure.
3. 2 percent: This is only the error in the side length \( L \) itself, without accounting for the separate error contributed by the force measurement, so it understates the total uncertainty in \( P \).
4. 8 percent: This would overstate the combined uncertainty for this measurement, since it does not correspond to simply combining the two given individual percentage errors of \( F \) and \( L \) as reported for this instrument.

Adding the two contributing percentage errors, one from the force reading and one carried through from the length reading, gives a combined uncertainty of 6% in the pressure.

Therefore, the correct answer is 6 percent.